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4 votes
1 answer
727 views

Adjunction between topological spaces and condensed sets

I am trying to prove that the functor \begin{align*} \mathrm{Top} &\longrightarrow \mathrm{Cond}(\mathrm{Set}) \\ X &\longmapsto \underline{X} \end{align*} admits a left adjoint and it is the ...
Luiz Felipe Garcia's user avatar
0 votes
1 answer
161 views

Sober spaces vs. spatial frames-a big picture

For any topological space $X$ one can consider the so called frame of all open subsets of $X$ to be denoted by $\mathcal{O}(X)$. If $f:X \to Y$ is continuous taking the inverse image we get the ...
truebaran's user avatar
  • 9,170
5 votes
0 answers
156 views

For which topological spaces does pullback along $\operatorname{ev}_0:B^I\to B$ have a right adjoint?

Let $B$ be a topological space. Consider the evaluation at zero of paths in $B$. This is a continuous map $\operatorname{ev}_0:B^I\to B$ where the domain carries the compact-open topology. For which ...
Arrow's user avatar
  • 10.3k
3 votes
1 answer
263 views

Adjoints for the functor ${\bf Top}\to {\bf Conv}$

Let $X$ be a set and let $\Phi(X)$ denote the collection of filters on $X$. For $x\in X$ we denote by $P_x$ the filter $P_x=\{A\subseteq X:x\in A\}$. A convergence space is a pair $(X,\to)$, where $X$ ...
Dominic van der Zypen's user avatar
14 votes
3 answers
688 views

Is there a monad on Set whose algebras are Tychonoff spaces?

Compact Hausdorff spaces are algebras of the ultrafilter monad on Set. Is the category of Tychonoff spaces also monadic over Set?
Gerrit Begher's user avatar
4 votes
1 answer
2k views

How is the right adjoint $f_*$ to the inverse image functor $f^*$ described for functor categories $Set^C$, $Set^D$ and $f : C \to D$

For $C,D$ small categories, and $f : C \to D$ a functor between them, there is a precomposition, or "inverse image", functor $f^* = (-) \circ f : Set^D \to Set^C$. It has a left and a right adjoint. ...
vincenzoml's user avatar